What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? (Weights can be placed only on the left pan.)

To weigh all the numbers by placing weight only on the left pan, we have to choose weights in increasing powers of $$2$$ starting from $$2^0$$.

Also, we need as many weights in increasing power till the sum of all weights is just greater than 100 (since we need minimum number of weights)

Let say $$n$$ weights be required, so the last weight will be of value $$2^{n-1}$$

Now, sum of all weights = $$2^0+2^1+2^2+2^3+.........+2^{n-1}=2^n-1$$

So, $$2^n-1>100$$

or, $$2^n>101$$

or, $$n=7$$ (as $$2^7=128$$ is just greater than $$101$$)

so, minimum number of weights required are $$7$$